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Exponential revision question I can't work out?

  1. Nov 6, 2014 #1
    < Mentor Note -- Thread moved from the Calculus forum to HH/Calculus >

    Hey everyone, I've started my new course and have been given revision and tutorial questions to do at home, this question has been bugging me for days now! I need to make "t" the subject to allow me to progress into the solution further.

    Make t the subject in " i=12.5(1-e^-t/cr) "
    I know that i need to use natural logs (ln) however i cannot seem to transpose the equation correctly. Could you please help me transpose this.

    Thanks
    Ramjam

    p.s forgot to add my working so far, DUH!
    ive transposed it to this form so far
    i/12.5 = 1 - e ^-t/cr

    this is where the logs come in and i have trouble, im not sure what to do next?!?
     
    Last edited by a moderator: Nov 6, 2014
  2. jcsd
  3. Nov 6, 2014 #2

    pasmith

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    Before taking logs, you must make [itex]e^{-t/cr}[/itex] the subject.
     
  4. Nov 6, 2014 #3

    Mark44

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    This is very unclear.
    The quantity in parentheses could be any of the following:
    ##1 - \frac{e^{-t}}{cr}##
    ##1 - \frac{e^{-t}}{c}r##
    ##1 - e^{-t/c * r}##
    No, logs don't come in yet. You still have that 1 on the right side.
     
  5. Nov 6, 2014 #4
    How do I go about removing the one on the right hand side, is it correct for me to move it to the left hand side, leaving e^-t/cr on the right?
     
  6. Nov 6, 2014 #5

    Ray Vickson

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    He might even mean ##\frac{1-e^{-t}}{cr}##. For some unknown reason, he refuses to deal with your question.
     
  7. Nov 6, 2014 #6
    My lecturer hinted that the 1 should be moved leaving e^t/cr on the right, however unsure how to transpose it to that
     
  8. Nov 6, 2014 #7

    vela

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    I moved this to the pre-calc forum since the issue is apparently about basic algebra.
     
  9. Nov 6, 2014 #8

    Mark44

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    Please address the question I asked in post #3.
     
  10. Nov 6, 2014 #9
    1-e-t/cr sorry my phone cut your post so I didn't understand what you meant
     
  11. Nov 6, 2014 #10

    Mark44

    Staff: Mentor

    You have
    What can you do to both sides of this equation so that the 1 on the right side is gone? Possible choices are to add the same number toboth sides, subtract the same number from both sides, multiply both sides by the same nonzero number, or divide both sides by the same nonzero number.
     
  12. Nov 6, 2014 #11
    If I was to add 1 to both sides e.g i/12.5 + 1 = 1-1e-t/cr is this acceptable? As then the one can be removed from the right ?
     
  13. Nov 6, 2014 #12

    Mark44

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    No, plus it's wrong.
    i/12.5 = 1 - e-t/cr

    If you add 1 to both sides, you get
    i/12.5 + 1 = 1 - e-t/cr + 1 = 2 - e-t/cr
    So that's no help at all.

    Note that 1e-t/cr is exactly the same as e-t/cr, so there's no point in writing a coefficient of 1 here.

    Nope.
     
  14. Nov 6, 2014 #13
    Could you possibly take me through step by step how to remove 1 from the right leaving e-t/cr so that I can see how it's done, as confusing my self now

    Ramjam
     
  15. Nov 6, 2014 #14

    Mark44

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    There is no "step by step" here - this is not a complicated process. Earlier in this thread you said,
    You added 1 to the left side, but you didn't add 1 to the right side. Whatever you do to one side of the equation, you have to do the same to the other side.

    If you have i/12.5 = 1 - e-t/cr, what can you add to the 1 on the right side to get rid of it? In other words, what + 1 = 0?
     
  16. Nov 6, 2014 #15
    Okay so adding -1 to both sides i/12.5 -1 = 1-e-t/cr -1
    which when becomes i/12.5 -1 = -e-t/cr

    isnt the minus infront the the "e" become an issue? as you can log "-e"

    Thankyou for your help on the matter
     
  17. Nov 6, 2014 #16

    mfb

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    You can still multiply both sides by the same factor...
    (and you should try those basic steps yourself)
     
  18. Nov 6, 2014 #17

    Mark44

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    Yes.
    So what can you do to get rid of the minus sign? Remember to do the same thing to both sides of the equation.
     
  19. Nov 6, 2014 #18
    Umm..... Im not actually to sure? :/
     
  20. Nov 6, 2014 #19

    mfb

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    Then try different things. At least one of them will work.
     
  21. Nov 6, 2014 #20
    Well ive been trying things for the past half hour with no luck, im not seeing how i can multiply both sides to remove the minus from the "e"
     
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